Question

Three vectors \(\vec a,\vec b,\vec c\) satisfy the condition \(\vec a+\vec b+\vec c=0\). If \(|\vec a|=3,|\vec b|=4\) and \(|\vec c|=2\), then find the value of \(\mu=\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a\).

Hint:

Square the condition a+b+c=0 and expand using the dot product.

Answer 1

Step 1: Key Approach:
Since \(\vec a+\vec b+\vec c=0\), we have \(|\vec a+\vec b+\vec c|^{2}=0\).

Step 2: Expanding the square:
\(|\vec a+\vec b+\vec c|^{2}=|\vec a|^{2}+|\vec b|^{2}+|\vec c|^{2}+2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)=0\).

Step 3: Substituting the magnitudes:
\(3^{2}+4^{2}+2^{2}+2\mu=0\ \Rightarrow\ 9+16+4+2\mu=0\ \Rightarrow\ 29+2\mu=0\).

Final Answer:
\(\mu=-\dfrac{29}{2}\).\[ \boxed{\mu=-\dfrac{29}{2}} \]